Chapter 7, Problem 67CP

(a)

To determine

The value of $x$ i.e. extension of the spring.

Answer to Problem 67CP

The value of $x$ is $\frac{3.62m}{4.30-23.4m}$.

Explanation of Solution

When the puck is set into circular motion then force applied by the spring must be equal to centripetal force to keep the system rotating.

Write the expression for force applied by the spring.

    $F=kx$                                                                                                            (I)

Here, $F$ is the force applied by the spring, $k$ is spring constant and $x$ is compression in the spring.

Write the expression for centripetal acceleration.

    $a_c={\omega }^{2}r$

Here, $a_c$ is centripetal acceleration, $\omega$ is angular speed, and r is distance at which object is kept during circular motion.

Write the expression for r.

    $r=x_{us}+x$

Here, $x_{us}$ is unstressed distance of the spring.

Write the expression for centripetal force.

    $F_c=m{a}_c$                                                                                                       (II)

Here, $F_c$ is centripetal force and $m$ is the mass.

Substitute ${\omega }^{2}r$ for ${\alpha }_c$ in equation (II).

    $F_c=m{\omega }^{2}r$                                                                                                  (III)

Write the expression for $\omega$ .

    $\omega =\frac{2\pi }{t}$

Here, $t$ is time period for which the object is in motion.

Substitute $\frac{2\pi }{t}$ for $\omega$ and $x_{us}+x$ for $r$ in equation (III).

    $F_c=m{\left(\frac{2\pi }{t}\right)}^2{x}_{us}+x$                                                                                   (IV)

Write the expression for conservation of forces.

    $F=F_c$

Substitute $m{\left(\frac{2\pi }{t}\right)}^2{x}_{us}+x$ for $F_c$ and $kx$ for $F$ in above equation.

    $kx=m{\left(\frac{2\pi }{t}\right)}^2\left(x_{us}+x\right)$                                                                                 (V)

Substitute $15.5\text{cm}$ for $x_{us}$, $1.30\text{s}$ for t, $4.30\text{N/m}$ for $k$ in equation (V).

    $\left(4.30\text{N/m}\right)x=m{\left(\frac{2\pi }{1.30\text{s}}\right)}^2\left(\left(15.5\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)+x\right)$

Simplify above equation.

    $\begin{array}{c}\left(4.30\right)x=m\left(23.36\right)\left(0.155\text{m}+x\right)\\ 3.62m=\left(4.30\right)x-23.36mx\end{array}$

Rearrange above equation for $x$ .

    $\begin{array}{c}x=\frac{3.62m}{4.30-23.36m}\\ \simeq \frac{3.62m}{4.30-23.4m}\end{array}$                                                                                      (VI)

Conclusion:

Thus, the value of $x$ is $\frac{\left(3.62\text{m}\right)m}{4.30\text{kg}-\left(23.4\right)m}$.

(b)

To determine

The value of $x$ for mass.

Answer to Problem 67CP

The value of $x$ for mass $0.070\text{kg}$ is $0.0951\text{m}$.

Explanation of Solution

Conclusion:

Substitute $0.070\text{kg}$ for $m$ in equation (VI).

    $\begin{array}{c}x=\frac{\left(3.62\text{ m}\right)\left(0.070\text{kg}\right)}{4.30\text{ kg}-23.4\left(0.070\text{kg}\right)}\\ =0.0951\text{m}\end{array}$

Thus, the value of $x$ for mass $0.070\text{kg}$ is $0.0951\text{m}$.

(c)

To determine

The value of $x$ for mass $0.140\text{kg}$.

Answer to Problem 67CP

The value of $x$ for mass $0.140\text{kg}$ is $0.492\text{m}$ .

Explanation of Solution

Conclusion:

Substitute $0.140\text{kg}$ for $m$ in equation (VI).

    $\begin{array}{c}x=\frac{\left(3.62\text{ m}\right)\left(0.140\text{kg}\right)}{4.30\text{ kg}-23.36\left(0.140\text{kg}\right)}\\ =0.492\text{m}\end{array}$

Thus, the value of $x$ for mass $0.140\text{kg}$ is $0.492\text{m}$.

(d)

To determine

The value of $x$ for mass $0.180\text{kg}$.

Answer to Problem 67CP

The value of $x$ for mass $0.180\text{kg}$ is $6.85\text{m}$.

Explanation of Solution

Conclusion:

Substitute $0.180\text{kg}$ for $m$ in equation (VI).

    $\begin{array}{c}x=\frac{\left(3.62\text{ m}\right)\left(0.180\text{kg}\right)}{4.30\text{ kg}-23.36\left(0.180\text{kg}\right)}\\ =6.8445\text{m}\\ \simeq \text{6}\text{.85}\text{m}\end{array}$

Thus, The value of $x$ for mass $0.180\text{kg}$ is $6.85\text{m}$.

(e)

To determine

The value of $x$ for mass $0.190\text{kg}$.

Answer to Problem 67CP

The value of $x$ for mass $0.190\text{kg}$ is $-4.97\text{m}$ which is impossible in this case as the value is negative and this is beyond infinity.

Explanation of Solution

Conclusion:

Substitute $0.190\text{kg}$ for $m$ in equation (VI).

    $\begin{array}{c}x=\frac{\left(3.62\text{ m}\right)\left(0.190\text{kg}\right)}{4.30\text{ kg}-23.36\left(0.190\text{kg}\right)}\\ =-4.97\text{m}\end{array}$

Thus, the value of $x$ for mass $0.190\text{kg}$ is $-4.97\text{m}$ which is impossible in this case as the value is negative and this is beyond infinity.

(f)

To determine

Pattern of variation of $x$ for $m$.

Answer to Problem 67CP

The extension of the string $x$ increases with $m$ till $m=0.184\text{kg}$ beyond this value, extension is infinity.

Explanation of Solution

The extension of the string $x$ is directly proportional to mass when mass is only in few grams so when mass increases, extension grows faster till $m=0.184\text{kg}$ as at this value extension becomes infinity.

When the denominator of the fraction goes to zero, then value of $x$ becomes infinite.

    $4.30-23.4m=0$

Solve above equation for $m$ .

    $m=0.184\text{kg}$

Therefore for $m=0.184\text{kg}$ value of extension is infinity so beyond this, situation not possible.

Conclusion:

Thus, the extension of the string $x$ increases with $m$ till $m=0.184\text{kg}$ beyond this value , extension is infinity.